Inequality involving square roots

**Ruun** · June 28th 2010, 11:04 AM

Hi, I've found a proof from $|x| \leq \sum_{i=1}^{n} |x_{i}|$ given $x=(x_{1},...,x_{n}), n \in \mathbb{N}$ It reads:

$\sum_{i=1}^{n}(x_{i})^{2} \leq \sum_{i=1}^{n} (x_{i})^2 + 2\sum_{i \ne j}^{n} |x_{i}||x_{j}| = (\sum_{i=1}^{n}|x_{i}|)^2$ .

And taking the square root it's supposed to grant the result.

Now, I do not see why $a^2 \geq b^2 \Rightarrow a \geq b$

Thanks in advance

**gmatt** · June 28th 2010, 11:12 AM

The statement $a^2 \ge b^2 \implies a \ge b$ is equivalent to $a < b \implies a^2 < b^2$ since they are contrapositives. Is it clear to you that $a < b \implies a^2 < b^2$ ?

**Ruun** · June 28th 2010, 11:19 AM

Given $a \leq b \Rightarrow a - b \leq 0$ Now I need information for $a+b$ because $(a+b)(a-b) = a^2-b^2$ , and that's the piece that I need.

I'm sorry but I still don't get it. Thanks for your post.

**Ackbeet** · June 28th 2010, 11:45 AM

Looking at this idea from a calculus perspective: the square root function is everywhere increasing and positive on its domain. Therefore, it preserves order.

**gmatt** · June 28th 2010, 11:47 AM

Well what you have written is fine. Given $a < b$ we have that $a-b < 0$ now provided $a + b \ge 0$ we have that $(a^2 -b^2) = (a+b)(a-b) < (a+b) 0 = 0.$

**Plato** · June 28th 2010, 12:29 PM

Originally Posted by Ruun

Now, I do not see why $a^2 \geq b^2 \Rightarrow a \geq b$

I do not understand what exactly you are asking for.
However, $a^2 \geq b^2 \Rightarrow a \geq b$ is clearly false.
It is true that $a^2 \geq b^2 \Rightarrow |a| \geq |b|$ .

Is it also the case that $\left| x \right| = \sqrt {\sum\limits_{k = 1}^n {\left( {x_k } \right)^2 } } ~?$
If so, the proof seems to work.

**Ruun** · June 28th 2010, 04:43 PM

Yes, it is the case.

Well the proof says taking square root, and I didn't find an argument to believe in that, and this is the point I'm looking for,

**Ackbeet** · June 28th 2010, 06:09 PM

Reply to Plato @ Post #6:

In this case, the things we're taking the square roots of are all positive. The step in question is this one:

$\sum_{i=1}^{n}(x_{i})^{2} \leq (\sum_{i=1}^{n}|x_{i}|)^2$

implies

$\sqrt{\sum_{i=1}^{n}(x_{i})^{2}} \leq \sum_{i=1}^{n}|x_{i}|.$

It is also true that $|x|=\sqrt{\sum_{i=1}^{n}(x_{i})^{2}}$ , at least in the Euclidean or $L^{2}$ norm.

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